{"id":872,"date":"2025-06-20T17:19:14","date_gmt":"2025-06-20T17:19:14","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=872"},"modified":"2025-09-09T19:53:00","modified_gmt":"2025-09-09T19:53:00","slug":"direction-fields-and-eulers-method-learn-it-2-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/direction-fields-and-eulers-method-learn-it-2-2\/","title":{"raw":"Introduction to Series: Learn It 2","rendered":"Introduction to Series: Learn It 2"},"content":{"raw":"

The Harmonic Series<\/h2>\r\n

One of the most important and surprising series in mathematics is the harmonic series<\/strong>:<\/p>\r\n\r\n

[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}=1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\cdots [\/latex].<\/div>\r\n

This series is fascinating because it diverges<\/strong>, but it does so extremely slowly. The partial sums grow toward infinity, but at such a slow rate that it's not immediately obvious the series diverges.<\/p>\r\n

Look at how slowly the partial sums [latex]S_k[\/latex] grow:<\/p>\r\n\r\n\r\n\r\n\r\n\r\n
[latex]k[\/latex]<\/td>\r\n[latex]10[\/latex]<\/td>\r\n[latex]100[\/latex]<\/td>\r\n[latex]1000[\/latex]<\/td>\r\n[latex]10,000[\/latex]<\/td>\r\n[latex]100,000[\/latex]<\/td>\r\n[latex]1,000,000[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]{S}_{k}[\/latex]<\/td>\r\n[latex]2.92897[\/latex]<\/td>\r\n[latex]5.18738[\/latex]<\/td>\r\n[latex]7.48547[\/latex]<\/td>\r\n[latex]9.78761[\/latex]<\/td>\r\n[latex]12.09015[\/latex]<\/td>\r\n[latex]14.39273[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nEven after adding one million terms, the partial sum is only about [latex]14.4[\/latex]. From this table alone, you might think the series converges to some finite value! Despite the slow growth, we can prove analytically that the harmonic series diverges by showing the partial sums are unbounded.\r\n\r\n
To show that the sequence of partial sums diverges, we show that the sequence of partial sums is unbounded.\r\n

Step 1: Group terms strategically<\/strong><\/p>\r\n

Let's examine the first few partial sums:<\/p>\r\n\r\n

    \r\n \t
  • [latex]S_1 = 1[\/latex]<\/li>\r\n \t
  • [latex]S_2 = 1 + \\frac{1}{2}[\/latex]<\/li>\r\n \t
  • [latex]S_4 = 1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4}[\/latex]<\/li>\r\n<\/ul>\r\n

    Step 2: Use inequalities to find lower bounds<\/strong><\/p>\r\n

    For [latex]S_4[\/latex], notice that: [latex]\\frac{1}{3} + \\frac{1}{4} > \\frac{1}{4} + \\frac{1}{4} = \\frac{1}{2}[\/latex]<\/p>\r\n

    Therefore, we conclude that: [latex]S_4 > 1 + \\frac{1}{2} + \\frac{1}{2} = 1 + 2 \\cdot \\frac{1}{2}[\/latex]<\/p>\r\n

    Step 3: Extend the pattern<\/strong><\/p>\r\n

    For [latex]S_8[\/latex]: [latex]S_8 = 1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5} + \\frac{1}{6} + \\frac{1}{7} + \\frac{1}{8}[\/latex]<\/p>\r\n

    Group the terms: [latex]S_8 > 1 + \\frac{1}{2} + \\left(\\frac{1}{4} + \\frac{1}{4}\\right) + \\left(\\frac{1}{8} + \\frac{1}{8} + \\frac{1}{8} + \\frac{1}{8}\\right)[\/latex]<\/p>\r\n

    [latex]S_8 > 1 + \\frac{1}{2} + \\frac{1}{2} + \\frac{1}{2} = 1 + 3 \\cdot \\frac{1}{2}[\/latex]<\/p>\r\n

    Step 4: Generalize the result<\/strong><\/p>\r\n

    Following this pattern:<\/p>\r\n\r\n

      \r\n \t
    • [latex]S_1 = 1[\/latex]<\/li>\r\n \t
    • [latex]S_2 = 1 + \\frac{1}{2}[\/latex]<\/li>\r\n \t
    • [latex]S_4 > 1 + 2 \\cdot \\frac{1}{2}[\/latex]<\/li>\r\n \t
    • [latex]S_8 > 1 + 3 \\cdot \\frac{1}{2}[\/latex]<\/li>\r\n<\/ul>\r\n

      In general: [latex]S_{2^j} > 1 + j \\cdot \\frac{1}{2}[\/latex] for all [latex]j \\geq 1[\/latex]<\/p>\r\n

      Step 5: Conclude divergence<\/strong><\/p>\r\n

      Since [latex]1 + j \\cdot \\frac{1}{2} \\to \\infty[\/latex] as [latex]j \\to \\infty[\/latex], the sequence [latex]{S_k}[\/latex] is unbounded and therefore diverges.<\/p>\r\n\r\n<\/section>

      \r\n

      harmonic series<\/h3>\r\nThe harmonic series [latex]\\sum_{n=1}^{\\infty} \\frac{1}{n} = 1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\cdots[\/latex] diverges, even though its terms approach zero.\r\n\r\n[latex]\\\\[\/latex]\r\n\r\nThis series demonstrates that having [latex]\\lim_{n \\to \\infty} a_n = 0[\/latex] is necessary but not sufficient for series convergence. The harmonic series diverges so slowly that its partial sums grow approximately like [latex]\\ln(n)[\/latex], making it a borderline case between convergence and divergence.\r\n\r\n<\/section>
      Why This Matters<\/strong>: The harmonic series shows that even when terms approach zero (like [latex]\\frac{1}{n} \\to 0[\/latex]), the series may still diverge. The rate at which terms approach zero matters greatly for convergence.<\/section>
      \r\n

      Algebraic Properties of Convergent Series<\/h2>\r\n

      Since infinite series are defined using limits of sequences, they inherit many algebraic properties from limits and sequences.<\/p>\r\n\r\n

      \r\n

      algebraic properties of convergent series<\/h3>\r\n

      Let [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] be convergent series. Then:<\/p>\r\n\r\n

        \r\n \t
      1. Sum Rule<\/strong>: The series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}+{b}_{n}\\right)[\/latex] converges and [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}+{b}_{n}\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}+\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex].<\/li>\r\n \t
      2. Difference Rule<\/strong>:The series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}-{b}_{n}\\right)[\/latex] converges and [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}-{b}_{n}\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}-\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex].<\/li>\r\n \t
      3. Constant Multiple Rule<\/strong>:For any real number [latex]c[\/latex], the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }c{a}_{n}[\/latex] converges and [latex]\\displaystyle\\sum _{n=1}^{\\infty }c{a}_{n}=c\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>
        \r\n
        \r\n

        Evaluate<\/p>\r\n\r\n

        [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[\\frac{3}{n\\left(n+1\\right)}+{\\left(\\frac{1}{2}\\right)}^{n - 2}\\right][\/latex].<\/div>\r\n \r\n\r\n<\/div>\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n
        \r\n

        We showed earlier that<\/p>\r\n\r\n

        [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n\\left(n+1\\right)}[\/latex]<\/div>\r\n \r\n

        and<\/p>\r\n\r\n

        [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{2}\\right)}^{n - 1}=2[\/latex].<\/div>\r\n \r\n

        Since both of those series converge, we can apply the Alegraic Properties of Convergent Series to evaluate<\/p>\r\n\r\n

        [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[\\frac{3}{n\\left(n+1\\right)}+{\\left(\\frac{1}{2}\\right)}^{n - 2}\\right][\/latex].<\/div>\r\n \r\n

        Using the sum rule, write<\/p>\r\n\r\n

        [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[\\frac{3}{n\\left(n+1\\right)}+{\\left(\\frac{1}{2}\\right)}^{n - 2}\\right]=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{3}{n\\left(n+1\\right)}\\underset{n=1}{\\overset{\\infty }{+\\displaystyle\\sum }}{\\left(\\frac{1}{2}\\right)}^{n - 2}[\/latex].<\/div>\r\n \r\n

        Then, using the constant multiple rule and the sums above, we can conclude that<\/p>\r\n\r\n

        [latex]\\begin{array}{ll}{\\displaystyle\\sum _{n=1}^{\\infty }\\frac{3}{n\\left(n+1\\right)}+\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{2}\\right)}^{n - 2}}\\hfill & =3{\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n\\left(n+1\\right)}+{\\left(\\frac{1}{2}\\right)}^{-1}\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{2}\\right)}^{n - 1}}\\hfill \\\\ & =3\\left(1\\right)+{\\left(\\frac{1}{2}\\right)}^{-1}\\left(2\\right)=3+2\\left(2\\right)=7.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><\/section>","rendered":"

        The Harmonic Series<\/h2>\n

        One of the most important and surprising series in mathematics is the harmonic series<\/strong>:<\/p>\n

        [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}=1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\cdots[\/latex].<\/div>\n

        This series is fascinating because it diverges<\/strong>, but it does so extremely slowly. The partial sums grow toward infinity, but at such a slow rate that it’s not immediately obvious the series diverges.<\/p>\n

        Look at how slowly the partial sums [latex]S_k[\/latex] grow:<\/p>\n\n\n\n\n
        [latex]k[\/latex]<\/td>\n[latex]10[\/latex]<\/td>\n[latex]100[\/latex]<\/td>\n[latex]1000[\/latex]<\/td>\n[latex]10,000[\/latex]<\/td>\n[latex]100,000[\/latex]<\/td>\n[latex]1,000,000[\/latex]<\/td>\n<\/tr>\n
        [latex]{S}_{k}[\/latex]<\/td>\n[latex]2.92897[\/latex]<\/td>\n[latex]5.18738[\/latex]<\/td>\n[latex]7.48547[\/latex]<\/td>\n[latex]9.78761[\/latex]<\/td>\n[latex]12.09015[\/latex]<\/td>\n[latex]14.39273[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

        Even after adding one million terms, the partial sum is only about [latex]14.4[\/latex]. From this table alone, you might think the series converges to some finite value! Despite the slow growth, we can prove analytically that the harmonic series diverges by showing the partial sums are unbounded.<\/p>\n

        To show that the sequence of partial sums diverges, we show that the sequence of partial sums is unbounded.<\/p>\n

        Step 1: Group terms strategically<\/strong><\/p>\n

        Let’s examine the first few partial sums:<\/p>\n

          \n
        • [latex]S_1 = 1[\/latex]<\/li>\n
        • [latex]S_2 = 1 + \\frac{1}{2}[\/latex]<\/li>\n
        • [latex]S_4 = 1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4}[\/latex]<\/li>\n<\/ul>\n

          Step 2: Use inequalities to find lower bounds<\/strong><\/p>\n

          For [latex]S_4[\/latex], notice that: [latex]\\frac{1}{3} + \\frac{1}{4} > \\frac{1}{4} + \\frac{1}{4} = \\frac{1}{2}[\/latex]<\/p>\n

          Therefore, we conclude that: [latex]S_4 > 1 + \\frac{1}{2} + \\frac{1}{2} = 1 + 2 \\cdot \\frac{1}{2}[\/latex]<\/p>\n

          Step 3: Extend the pattern<\/strong><\/p>\n

          For [latex]S_8[\/latex]: [latex]S_8 = 1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5} + \\frac{1}{6} + \\frac{1}{7} + \\frac{1}{8}[\/latex]<\/p>\n

          Group the terms: [latex]S_8 > 1 + \\frac{1}{2} + \\left(\\frac{1}{4} + \\frac{1}{4}\\right) + \\left(\\frac{1}{8} + \\frac{1}{8} + \\frac{1}{8} + \\frac{1}{8}\\right)[\/latex]<\/p>\n

          [latex]S_8 > 1 + \\frac{1}{2} + \\frac{1}{2} + \\frac{1}{2} = 1 + 3 \\cdot \\frac{1}{2}[\/latex]<\/p>\n

          Step 4: Generalize the result<\/strong><\/p>\n

          Following this pattern:<\/p>\n

            \n
          • [latex]S_1 = 1[\/latex]<\/li>\n
          • [latex]S_2 = 1 + \\frac{1}{2}[\/latex]<\/li>\n
          • [latex]S_4 > 1 + 2 \\cdot \\frac{1}{2}[\/latex]<\/li>\n
          • [latex]S_8 > 1 + 3 \\cdot \\frac{1}{2}[\/latex]<\/li>\n<\/ul>\n

            In general: [latex]S_{2^j} > 1 + j \\cdot \\frac{1}{2}[\/latex] for all [latex]j \\geq 1[\/latex]<\/p>\n

            Step 5: Conclude divergence<\/strong><\/p>\n

            Since [latex]1 + j \\cdot \\frac{1}{2} \\to \\infty[\/latex] as [latex]j \\to \\infty[\/latex], the sequence [latex]{S_k}[\/latex] is unbounded and therefore diverges.<\/p>\n<\/section>\n

            \n

            harmonic series<\/h3>\n

            The harmonic series [latex]\\sum_{n=1}^{\\infty} \\frac{1}{n} = 1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\cdots[\/latex] diverges, even though its terms approach zero.<\/p>\n

            [latex]\\\\[\/latex]<\/p>\n

            This series demonstrates that having [latex]\\lim_{n \\to \\infty} a_n = 0[\/latex] is necessary but not sufficient for series convergence. The harmonic series diverges so slowly that its partial sums grow approximately like [latex]\\ln(n)[\/latex], making it a borderline case between convergence and divergence.<\/p>\n<\/section>\n

            Why This Matters<\/strong>: The harmonic series shows that even when terms approach zero (like [latex]\\frac{1}{n} \\to 0[\/latex]), the series may still diverge. The rate at which terms approach zero matters greatly for convergence.<\/section>\n
            \n

            Algebraic Properties of Convergent Series<\/h2>\n

            Since infinite series are defined using limits of sequences, they inherit many algebraic properties from limits and sequences.<\/p>\n

            \n

            algebraic properties of convergent series<\/h3>\n

            Let [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] be convergent series. Then:<\/p>\n

              \n
            1. Sum Rule<\/strong>: The series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}+{b}_{n}\\right)[\/latex] converges and [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}+{b}_{n}\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}+\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex].<\/li>\n
            2. Difference Rule<\/strong>:The series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}-{b}_{n}\\right)[\/latex] converges and [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}-{b}_{n}\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}-\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex].<\/li>\n
            3. Constant Multiple Rule<\/strong>:For any real number [latex]c[\/latex], the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }c{a}_{n}[\/latex] converges and [latex]\\displaystyle\\sum _{n=1}^{\\infty }c{a}_{n}=c\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex].<\/li>\n<\/ol>\n<\/section>\n
              \n
              \n

              Evaluate<\/p>\n

              [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[\\frac{3}{n\\left(n+1\\right)}+{\\left(\\frac{1}{2}\\right)}^{n - 2}\\right][\/latex].<\/div>\n

               <\/p>\n<\/div>\n